4.2 State the principle of conservation of linear momentum in words.
This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Here are the solutions to the questions:
4.2 State the principle of conservation of linear momentum in words.
The principle of conservation of linear momentum states that the total linear momentum of an isolated system remains constant in both magnitude and direction. This means that in the absence of external forces, the total momentum before a collision or interaction is equal to the total momentum after the collision or interaction.
4.3 Calculate the velocity of the truck-car system immediately after the collision.
Step 1: Define the given values and choose a positive direction.
Let EAST be the positive direction.
Mass of car, $m_c = m$
Initial velocity of car, $v_{c,i} = +20 \text{ m/s}$
Mass of truck, $m_t = 2m$
Initial velocity of truck, $v_{t,i} = -20 \text{ m/s}$
Since they stick together, their final velocity will be the same, $v_f$.
Step 2: Apply the principle of conservation of linear momentum.
$$ m_c v_{c,i} + m_t v_{t,i} = (m_c + m_t) v_f $$
Step 3: Substitute the values into the equation.
$$ (m)(+20) + (2m)(-20) = (m + 2m) v_f $$
$$ 20m - 40m = (3m) v_f $$
$$ -20m = 3m v_f $$
Step 4: Solve for $v_f$.
$$ v_f = \frac{-20m}{3m} $$
$$ v_f = -\frac{20}{3} \text{ m/s} $$
$$ v_f \approx -6.67 \text{ m/s} $$
The negative sign indicates the direction is WEST.
The velocity of the truck-car system immediately after the collision is $\boxed{\text{6.67 m/s WEST}}$.
4.4 On impact the car exerts a force of magnitude $F$ on the truck and experiences an acceleration of magnitude $a$.
4.4.1 Determine, in terms of $F$, the magnitude of the force that the truck exerts on the car on impact. Give a reason for the answer.
The magnitude of the force that the truck exerts on the car is $\boxed{F}$.
Reason: According to Newton's Third Law of Motion, for every action, there is an equal and opposite reaction. The force exerted by the car on the truck is equal in magnitude and opposite in direction to the force exerted by the truck on the car.
4.4.2 Determine, in terms of $a$, the acceleration that the truck experiences on impact. Give a reason for the answer.
Step 1: Apply Newton's Second Law to the car.
The car has mass $m$ and experiences an acceleration of magnitude $a$.
The magnitude of the force exerted on the car by the truck is $F$ (from 4.4.1).
$$ F_{net,car} = m_c a_c $$
$$ F = m a $$
Step 2: Apply Newton's Second Law to the truck.
The truck has mass $2m$. Let its acceleration be $a_t$.
The magnitude of the force exerted on the truck by the car is $F$ (given in the question).
$$ F_{net,truck} = m_t a_t $$
$$ F = (2m) a_t $$
Step 3: Equate the expressions for $F$ and solve for $a_t$.
$$ m a = (2m) a_t $$
$$ a_t = \frac{ma}{2m} $$
$$ a_t = \frac{a}{2} $$
The acceleration that the truck experiences on impact is $\boxed{\frac{a}{2}}$.
Reason: According to Newton's Second Law of Motion ($F_{net} = ma$), for the same magnitude of force $F$, an object with twice the mass ($2m$) will experience half the acceleration ($a/2$) compared to an object with mass $m$.
4.4.3 Both drivers are wearing identical seat belts. Which driver is likely to be more severely injured on impact? Explain the answer by referring to acceleration and velocity.
The car driver is likely to be more severely injured.
Explanation:
• The car has a mass of $m$, and the truck has a mass of $2m$.
• From question 4.4.2, the car experiences an acceleration of magnitude $a$, while the truck experiences an acceleration of magnitude $\frac{a}{2}$. This means the car (and its driver) undergoes a greater deceleration during the collision.
• The car's initial velocity is $20 \text{ m/s}$ EAST, and its final velocity is $6.67 \text{ m/s}$ WEST. The magnitude of its change in velocity is $|(-6.67) - 20| = 26.67 \text{ m/s}$.
• The truck's initial velocity is $20 \text{ m/s}$ WEST, and its final velocity is $6.67 \text{ m/s}$ WEST. The magnitude of its change in velocity is $|(-6.67) - (-20)| = |-6.67 + 20| = 13.33 \text{ m/s}$.
• Since the car experiences a larger change in velocity over the same collision time, it also experiences a larger acceleration ($a = \frac{\Delta v}{\Delta t}$). A larger acceleration (deceleration) means a greater force exerted on the driver by the seatbelt, leading to more severe injuries.
3 done, 2 left today. You're making progress.
